# How to show $f(x) \leq 1+\frac{\pi}{4}$ for every $x \geq 1$

Suppose $f$ is a real-valued differentiable function defined on $[ 1,\infty)$ with $f(1)=1$. Suppose , moreover , that $f$ satisfies $$f'(x)=\frac{1}{x^2+f^2(x)}$$ Show that $f(x) \leq 1+\frac{\pi}{4}$ for every $x \geq 1$ .

Trial: I try to find the maximum value of $f(x)$ but here $f'(x)=0$ has no solution.

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Hint: $f(x)-f(1)=\int_1^x f'(t)\,dt\le\int_1^x{1\over t^2+1}\,dt$. –  David Mitra Jan 23 '13 at 13:39
as david said:From the given condition $f'(x)={1\over x^2+f^2(x)}$, it follows that $f'$ is positive on $[1,\infty)$. So $f$ is increasing on $[1,\infty)$, and since $f(1)=1$, we have $f(x)\ge1$ on $[1,\infty)$. The aforementioned condition also implies that $f'$ is continuous on $[1,\infty)$. We thus have $$f(t)-f(1)=\int_1^t {1\over x^2+f^2(x)}\,dx\le \int_1^\infty{1\over 1+x^2}\,dx=\tan^{-1} x\,\bigl|_1^\infty={\pi\over 4};$$ whence the result follows.